MCQ Questions for Class 10 Maths Chapter 10 Circles

Circles Class 10 MCQ Questions with Answers

Question 1

If the radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is the tangent to the other circle is:

(A) 3 cm
(B) 6 cm
(C) 9 cm
(D) 1cm
Answer:
(B) 6 cm

Explanation:
Let C₁; C₂ be two coni oniric circles with their centre C.

Chord AB of circle C² touches C₁ at P AB is tangent at P and PC is radius
CP ⊥ AB
Given, ∠P = 90°, CP = 4 cm and CA = 5 cm
∴ IN is tangent angle ∆PAC,
AP² = AC² – PC²
= 5² – 4²
= 25 – 16
= 9
AP = 3 cm
∴ Perpendicular from centre to chord bisects the chord.
∴ AB = 2AP
= 2 × 3
= 6 cm.

Question 2.

In the given figure, if ∠AOB = 125°, then ∠COD is equal to:

(A) 62.5°
(B) 45°
(C) 35°
(D) 55°

Answer:
(D) 55°

Explanation:
Since, quadrilateral circumscribing a circle subtends supplementary angles at the centre of the circle.
∴∠AOB + ∠COD = 180°
125° + ∠COD = 180°
∠COD = 180° – 125° = 55°

Question 3.

In the given figure, AB is a chord of the circle and AOC is its diameter, such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to:

(A) 65°
(B) 60°
(C) 50°
(D) 40°

Answer:
(C) 50°

Explanation:
Since, the angle between chord and tangent is equal to the angle subtended by the same chord in alternate segment of circle. ⇒ ∠BAT = 50°.

Question 4.

From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is:

(A) 60 cm²
(B) 65 cm²
(C) 30 cm²
(D) 32.5 cm²

Answer:
(A) 60 cm²

Explanation:
PQ is tangent and QO is radius at contact point Question
∴ PQO = OP² – OQ²
= 132 – 52
= 169 – 25 = 144
PQ = 12 cm
∆OPQ = ∆OPR [SSS congruence]
∴ Area of ∆OPQ = area of ∆OPR
[Since, congruent figures are equal in areas]
Area of quadrilateral QORP = 2 area of ∆OPR
= 2 × \(\frac{1}{2}\) base × height
= RP × OR
= 12 × 5
= 60 cm²

Question 5.

At one end A of diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the cirde. The length of the chord CD parallel to XY and at a distance 8 cm from A is:

(A) 4cm
(B) 5cm
(C) 6cm
(D) 8cm
Answer:
(D) 8cm

Explanation:
XAY is tangent and AO is radius at contact point A of circle.

AO = 5 cm
∠OAY = 90°
∴ CD is another chord at distance (perpendicular) of 8 cm from A and CMD || XAY meets AB at M.
join OD.
OD = 5 cm
OM = 8 – 5 = 3 cm
∠OMD = ∠OAY = 90°
Now, in right angled ∆OMD
MD2 = OD2 – MO2
= 52 – 32
= 25 – 9
= 16
⇒MD = 4 cmWe know thaat perpendiculars from centre O of circle bisect the chorde.
CD = 2MD
= 2 × 4
= 8 cm.
Hence, length of chord, CD = 8 cm.

Question 6.

In the given figure, AT is a tangent to the circle with centre ‘O’ such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to:

(A) 4 cm
(B) 2 cm
(C) 2√3 cm
(D) 4√3 cm

Answer:
(C) 2√3 cm

Explanation:
Join OA. OA is radius and AT is tangen at contact point A.
∴∠OAT = 90°,
Given that, OT = 4 cm
Now, \(\frac{A T}{4}\) = \(\vec{a}\) = cos 30°
⇒ AT = 4 × \(\frac{\sqrt{3}}{2}\) = 2√3 cm.

Question 7.

In the given figure, ‘O’ is the centre of circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then POQ is equal to:

(A) 100°
(B) 80°
(C) 90°
(D) 75°

Answer:
(A) 100°

Explanation:
OP is radius and PR is tangent at P.
so, ∠OPR = 90°
∠OPQ + 50 = 90°
∠OPQ = 90 – 50°
∠OPQ = 40°
In OPQ, OP = OQ
[Radii of same circle]
∴∠Q = ∠OPQ = 40°
[ Angle opposite to equal sides are equal]
But, ∠POQ = 180° – ∠P – ∠QO= 180° – 40° – 40°
= 180 – 80° = 100°
⇒ ∠POQ = 100° .

Question 8.

In the given figure, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to:

(A) 25°
(B) 30°
(C) 40°
(D) 50°

Answer:
(A) 25°

Explanation:
In ∆OAB, we have OA = OB [Radii of same circle ]
∴∠OAB = ∠OBA [Angle opposite to equal sides are equal]
As OA and PA are radius and tangent respectively at contact point A.
So, ∠OAP = 90°
Similariy, ∠OBP = 90°
Now, in quadrilateral PAOB,
∠P + ∠A + ∠O +∠B = 360°
⇒ 50° + 90 +∠ O + 90 = 360°
⇒ ∠O = 360° – 90° – 90° – 50°
⇒ ∠O = 130°
Again, in ∆OAB,
∠O + ∠OAB +∠ OBA = 180°
⇒ 130° +∠OAB + ∠OAB = 180° [∴ ∠OBA = ∠OAB]
⇒ 2 ∠OAB = 180° – 130° = 50v
⇒ ∠OAB = 25°
Hence,∠OAB = 25°

Assertion and Reason Based MCQs

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is True

Question 1.

Assertion (A): The length of the tangent drawn from a point 8 cm away from the centre of circle of radius 6 cm is 2 √7 cm.
Reason (R): If the angle between two radii of a circle is 130°, then the angle between the tangents at the end points of radii at their point of intersection is 50°.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In case of assertion:
Length of the tangent =\(\sqrt{d^{2}-r^{2}}\)
= \(\sqrt{(8)^{2}-(6)^{2}}\)
= \(\sqrt{64-36}\)
= \(\sqrt{28}\) = 2√7 cm
∴ Assertion is correct.
In case of reason:
Since, sum of the angles between radii and between intersection point of tangent is 180°. Angle at the point of intersection of tangents = 180° -130° = 50°.
∴ Reason is correct.
Hence, both assertion and reason are correct but reason is not the correct explanation for assertion.

Question 2.

Assertion (A): If a chord AB subtends an angle of 60° at the centre of a circle, then the angle between the tangents at A and B is also 60°.
Reason (R): The length of the tangent from an external point P on a circle with centre O is always less than OP.

Answer:
(D) A is false and R is True

Explanation:
In case of assertion:
Chord AB subtends ∠60° at O

∴ ∠OAP = 90°
Similarly, ∠OBP = 90°
In quadrilateral OAPB,
∠O + ∠P + ∠OAP + ∠OBP = 360°
⇒ 60 + ∠P + 90° +90° = 360°
⇒ ∠P = 360° – 240°
⇒ ∠P = 120°
∴ Assertion is correct.
In case of reason:
PT and OT are the tangent and radius, respectivly at contact point T.

So, OTP = 90°
⇒ ∆OPT is right angled triangle.
Again, in ∆OPT
∴ ∠T > ∠O
∴ ∠OP > ∠PT
[ side opposite to greater angle is larger]
∴ Reason is correct.
Hence, assertion is incorrect but reason is correct.

Question 3.

Assertion (A): The tangent to the circumcircle of an isosceles AABC at A, in which AB = AC, is parallel to BC.
Reason (R): PQ is a tangent drawn from an external point P to a circle with centre O. QOR is the diameter of the circle. If ∠POR = 120°, then the measure of ∠OPQ is 60°.

Answer:
(C) A is true but R is false

Explanation:
In case of assertion:
given that, ∆ABC, inscribed in a circle in which AB = AC

PAQ is tangent at A.
AB is chord.
∴∠ PAB = ∠C
As, we know that , ∠PAB formed by chord AB with tangente segment.
In ∆ABC,
AB = AC [ Given]
∴ ∠B = ∠C [∴Angles opposite to equai sides are equal]……(ii)
From (i) and (ii), ∠B = ∠PAB
These are alternate interior angles.
Therefore, PAQ || BC.
∴ Assertion is correct.
In case of reason:

Civen that,
PQ is a tangent, ∠POR = 120°
∠POR = ∠OQP + ∠OPQ [ Exterior angle sum property]
∠OPQ = 120 – 90° = 30°
∠OPQ = 30°
∴ Reason is correct.
Hence, assertion is correct but reason is incorrect.

Question 4.

Assertion (A): PQ is a tangent to a circle with centre O at point P. If AOPQ is an isosceles triangle, then ∠OQP = 45°.
Reason (R): If two tangents inclined at 60° are drawn to a circle of radius 3 cm, then the length of each tangent is 3√3 cm.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In case of assertion:
As we know that ∠OPQ = 90° ( Angle between tangent and radius)
Let ∠PQO be x°, then∠QOP = x°
Since∠ OPQ is an isosceles triangle.
(given)(OP = OQ)

In ∆OPQ,
∠OPQ + ∠PQO + ∠QOP = 180
(property of the sum of angles of a triangle)
∴ 90° + x° + x° = 180°
⇒ 2x° = 180° – 90° = 90°
⇒ x = \(\frac{90}{2}\) = 45.
Hence, OQP is 45

∴ Assertion is correct.
In case of reason:

In ∆PAO,
tan 30 = \(\frac{A O}{P A}\) (Using trigonometry)
\(\frac{1}{\sqrt{3}}\) = \(\frac{3}{P A}\)
PA = 3 √3 cm.
∴ Reason is correct.
Hence, assertion is correct but reason is incorrect.

Case – Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.
I. Read the following text and answer the following question on the basis of the same:
There is a circular filed of radius 5 m. Kanabh, Chikoo and Shubhi are playing with ball, in which Kanabh and Chikoo are standing on the boundary of the circle. The distance between Kanabh and Chikoo is 8 m. From Shubhi point S, two tangents are drawn as shown in the figure. Give the answer of the following questions.

Question 1.

What is the relation between the lengths of SK and SC?

(A) SK ≠ SC
(B) SK = SC
(C) SK > SC
(D) SK < SC.
Answer:
(B) SK = SC

Explanation:
We know that the lengths of tangents drawn from an external point to a circle are equal. So SK and SC are tangents to a circle with centre O.
∴SK = SC

Question 2.

The length (distance) of OR is:

(A) 3 m
(B) 4 m
(C) 5 m
(D) 6 m.
Answer:
(A) 3 m

Explanation:
In question 1, we have proved SK = SC
Then ASKC is an isosceles triangle and SO is the angle bisector of ZKSC. So, OS X KC.
∴ OS bisects KC, gives KR = RC = 4 cm
Now, OR = \(\sqrt{O K^{2}-K R^{2}}\)
(By using Pythagoras theorem)
= \(\sqrt{5^{2}-4^{2}}=\) = \(\sqrt{25-16}\)
= \(\sqrt{25-16}\)
= \(\sqrt{9}\)
= 3 m

Question 3.

The sum of angles SKR and OKR is:

(A) 45°
(B) 30°
(C) 90°
(D) none of these
Answer:
(C) 90°
∠SKR + ∠OKR = ∠OKR
= 90 (Radius is ⊥ to tangent)

Question 4.

The distance between Kanabh and Shubhi is:

(A) \(\frac{10}{3} \mathrm{~m}\)
(B) \(\frac{13}{3} \mathrm{~m}\)
(C) \(\frac{16}{3} \mathrm{~m}\)
(D) \(\frac{20}{3} \mathrm{~m}\)
Answer:
(D) \(\frac{20}{3} \mathrm{~m}\)

Explanation:
∆SER and ∆RKO,
∠RKO = ∠KSR
∠SRK = ∠ORK
∆KSR ~ ∆OKR . (By AA Similarity)
\(\frac{S K}{K O}\) = \(\frac{R K}{R O}\)
Then \(\frac{S K}{5}\) = \(\frac{4}{3}\)
(RO = 3 m, proved in Question 2)
⇒ 3SK = 20
⇒ Sk = \(\frac{20}{3}\)
Hence, the distance between Kanabh and Shubhi is \(\frac{20}{3}\)

Question 5.

What is the mathematical concept related to this question ?

(A) Constructions
(B) Area
(C) Circle
(D) none of these
Answer:
(C) Circle

Explanation:
The mathematical concept (Circle) is related to this question.

II. Read the following text and answer the following question on the basis of the same:
ABCD is a playground. Inside the playground a circular track is present such that it touches AB at point P, BC at Q, CD at R and DA at S.

Question 1.

If DR = 5 m, then DS is equal to:

(A) 6 m
(B) 11 m
(C) 5 m
(D) 18 m
Answer:
(C) 5 m

Explanation:
∴ DR =5 m (given)
DR = DS (Length of tangents are equal)
DS = 5 m.

Question 2.

The length of AS is:

(A) 18 m
(B) 13 m
(C) 14 m
(D) 12 m
Answer:
(A) 18 m

Explanation:
We have AD = 23 m.
ands DS = 5 m (Proved in Question 1)
∴ AS = AD – DS
= (23 – 5)m = 18 m.

Question 3.

The length of PB is:

(A) 12 m
(B) 11 m
(C) 13 m
(D) 20 m
Answer:
(B) 11 m

Explanation:
We have,
AB = 29 m
But AS = AP (lengths of tangents are equal)
and AS = 18 m(proved in Question 2)
∴ AP = 18 m
NOw, PB = AB – AP
= (29 – 18) m
= 11 m

Question 4.

What is the angle of OQB?

(A) 60°
(B) 30°
(C) 45°’
(D) 90°
Answer:
(D) 90°

Explanation:
∠OQB = 90°
(Radius is ’ to tangent)

Question 5.

What is the diameter of given circle?

(A) 22 m
(B) 33 m
(C) 20 m
(D) 30 m
Answer:
(A) 22 m

Explanation:
PB = 11 m (proved in Question 3) PB = BQ (lengths oftangents are equal)
But PB = BQ
∴BQ = 11 m
or = OQ = QB = 11m
Hence, diameter = 2r = 2 × 11 = 22 m.

III. Read the following text and answer the following question on the basis of the same:
A Ferris wheel (or a big wheel in the United Kingdom) is an amusement ride consisting of a rotating upright wheel with multiple passenger-carrying components (commonly referred to as passenger cars, cabins, tubs, capsules, gondolas, or pods) attached to the rim in such a way that as the wheel turns, they are kept upright, usually by gravity.After taking a ride in Ferris wheel, Aarti came out from the crowd and was observing her friends who were enjoying the ride. She was curious about the different angles and measures that the wheel will form. She forms the figure as given below.

Question 1.

In the given figure find ∠ROQuestion

(A) 60°
(B) 100°
(C) 150°
(D) 90°
Answer:
(C) 150°

Explanation:
∠ORP = 90° & ∠OQP [∵ radius of circle is perpendicular to tangent]
∴ ∠ROQ + ∠ORP + ∠OQP + ∠QPR = 360° ∠ROQ + 90° + 90° + 30° = 360° ∠ROQ + 210° = 360° ∠ROQ = 360°-210°
∠ROQ = 150°

Question 2.

Find ∠RQP.

(A) 75°
(B) 60°
(C) 30°
(D) 90°
Answer:
(A) 75°

Explanation:
∠OQR = ∠ORQ
∠RPQ = 150
and ∠ROQ + ∠OQR + ∠ORQ = 180
150 + 2∠ORQ =180
2∠ORQ = 30
∠ORQ = 15
∠OQR = ORQ = 1
∠RQP = ∠OQP – ∠OQR= 90 – 15

Question 3.

Find ∠RSQuestion

(A) 60°
(B) 75
(C) 100°
(D) 30
Answer:
(B) 75

Question 4.

Find ∠ORP

(A) 90°
(B) 70
(C) 100°
(D) 60
Answer:
(A) 90°

Explanation:
∠ORP = 90°
Because, radius of circle is perpendicular to tangent.

IV Read the following text and answer the following question on the basis of the same:
Varun has been selected by his School to design logo for Sports Day T-shirts for students and staff. The logo design is as given in the figure and he is working on the fonts and different colours according to the theme. In given figure, a circle with centre O is inscribed in a AABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. The lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively.

Question 1.

Find the length of AD.

(A) 7
(B) 8
(C) 5
(D) 9
Answer:
(A) 7

Explanation:
Given, AB = 12 cm
BC = 8 cm
CA = 10 cm

In ∆ABC
x + y =12 cm -(i)
x + z = 10 cm …(ii)
y + z = 8 cm -(Mi)
Adding (i), (ii) and (iii)
2(x + y + z) =30 cm
x + y + z =15 cm …(iv)
from eQuestion (iii) and (iv)
x = 7 cm
AD = x = 7 cm

Question 2.

Find the Length of BE.

(A) 8
(B) 5
(C) 2
(D) 9
Answer:
(B) 5

Explanation:
We have
x + y + z = 15 cm
and x + z = 10 cm
By solving above equations
we get y = 5 cm
i.e., BE = y = 5 cm

Question 3.

Find the length of CF.

(A) 9
(B) 5
(C) 2
(D) 3
Answer:
(D) 3

Explanation:
We have
x y + z = 15 cm
and x + y =12 cm
By solving above equations,
we get z = 3 cm
i.e., CF = z = 3 cm

Question 4.

If radius of the circle is 4 cm, find the area of ∆OAB.

(A) 20
(B) 36
(C) 24
(D) 48
Answer:
(C) 24

Explanation:
Area of ∆OAB = \(\frac{1}{2}\) × AB × OD
= \(\frac{1}{2}\) × 12 × 4
= 24 cm²

MCQ Questions for Class 10 Maths with Answers